- The point \(A\) has coordinates \((5,16)\) and the point \(B\) has coordinates \((-4,4)\). The variable point \(P\) has coordinates \((x,y)\) and moves on a path such that \(AP = 2BP\). Show that the Cartesian equation of the path of \(P\) is \[(x + 7)^2 + y^2 = 100.\]

We are told that \(AP = 2BP\), which is the same as \((AP)^2 = 4(BP)^2\).

We use the formula for the distance \(d\) between two points \((x_1,y_1)\) and \((x_2,y_2)\), which is \(d^2 = (x_2-x_1)^2 + (y_2-y_1)^2\).

This is the equation of a circle with centre \((-7,0)\) and radius \(10\).

- The point \(C\) has coordinates \((a,0)\) and the point \(D\) has coordinates \((b,0)\). The variable point \(Q\) moves on a path such that \[QC = k \times QD,\] where \(k > 1\). Given that the path of \(Q\) is the same as the path of \(P\), show that \[\frac{a + 7}{b + 7} = \frac{a^2 + 51}{b^2 + 51}.\] Show further that \((a + 7)(b + 7) = 100\), in the case \(a \neq b\).

If this locus is the same as the locus of \(P\), which had equation \(x^2+y^2+14x-51=0\), then we must be able to rewrite our equation in this form. (Alternatively, we can say that the ratios of corresponding coefficients must be the same.)

Note that \(k^2 - 1 \neq 0\), so we can divide by it to make the coefficients of the \(x^2\) and \(y^2\) terms equal to \(1\).

Dividing by \(k^2-1\) gives the locus of \(Q\) as \[x^2 + y^2 + \frac{2a - 2b k^2}{k^2 - 1}x + \frac{b^2 k^2 - a^2}{k^2 - 1} = 0.\]

If this locus is the same as the locus of \(P\), then we must have \[\frac{2a - 2b k^2}{k^2 - 1} = 14\] and \[\frac{b^2 k^2 - a^2}{k^2 - 1} = -51.\]

The first of these gives \[2a - 2b k^2 = 14k^2 - 14,\] so \[k^2 = \frac{a + 7}{b + 7}.\]

Similarly, the second rearranges to

\[b^2 k^2 - a^2 = -51k^2 + 51,\] so \[k^2 = \frac{a^2 + 51}{b^2 + 51}.\]

Thus \[\frac{a + 7}{b + 7} = \frac{a^2 + 51}{b^2 + 51}.\]

Now we can rearrange this to \[\begin{align*} &&(a + 7)(b^2 + 51) &= (b + 7)(a^2 + 51)&&\quad\\ \iff\quad&& ab^2 + 7b^2 + 51a &= a^2b + 7a^2 + 51b\\ \iff\quad&& ab^2 - a^2b &= 7(a^2 - b^2) + 51(b - a)\\ \iff\quad&& ab(b - a) &= 7(a - b)(a + b) + 51(b - a). \end{align*}\]Since \(a \neq b\), we may divide by \(b-a\) to obtain \[ab = 51 - 7(a + b).\]

That is, \[\begin{align*} &&ab + 7(a + b) &= 51&&\quad\\ \Longrightarrow \quad&& ab + 7a + 7b + 49 &= 51 + 49\\ \Longrightarrow \quad&& (a + 7)(b + 7) &= 100. \end{align*}\]